Question

我正在尝试使用超级库来发出一些请求。 Headers::get()方法返回var voipRegistry: PKPushRegistry! func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey : Any]? = nil) -> Bool { voipRegistry = PKPushRegistry(queue: DispatchQueue.main) voipRegistry.delegate = self voipRegistry.desiredPushTypes = Set([.voIP]) } extension AppDelegate: PKPushRegistryDelegate { func pushRegistry(_ registry: PKPushRegistry, didInvalidatePushTokenForType type: PKPushType) { print("didInvalidatePushTokenForType") } func pushRegistry(_ registry: PKPushRegistry, didReceiveIncomingPushWith payload: PKPushPayload, forType type: PKPushType) { print("Incoming voip notfication: \(payload.dictionaryPayload)") } func pushRegistry(_ registry: PKPushRegistry, didUpdate credentials: PKPushCredentials, forType type: PKPushType) { print("voip token: \(credentials.token)") } }，其中Option<&H>是一个带有一个字段的元组结构。我可以使用H来解构if let Some()。但是我们如何解构Option？当然，我总是可以使用&H访问该字段，但我很好奇Rust是否有语法来执行此操作。

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Answer 1

当您尝试编译它时，Rust编译器将告诉您如何使用一条好消息修复错误：

error[E0507]: cannot move out of borrowed content
  --> <anon>:11:9
   |
11 |     let &s(bar) = f(&my_struct2); // this does not work
   |         ^^^---^
   |         |  |
   |         |  hint: to prevent move, use `ref bar` or `ref mut bar`
   |         cannot move out of borrowed content

这需要告诉编译器你只需要引用struct中的字段;默认匹配将执行移动，原始struct值将不再有效。

让我们修复一下这个例子：

struct s(String);

fn f(input: &s) -> &s {
    input
}

fn main() {
    let my_struct1 = s("a".to_owned());
    let s(foo) = my_struct1;
    let my_struct2 = s("b".to_owned());
    let &s(ref bar) = f(&my_struct2);
}

另一种方法是先解除引用并删除&。我认为这在Rust中是首选：

struct s(String);

fn f(input: &s) -> &s {
    input
}

fn main() {
    let my_struct1 = s("a".to_owned());
    let s(foo) = my_struct1;
    let my_struct2 = s("b".to_owned());
    let s(ref bar) = *f(&my_struct2);
}

如何使用引用来构造元组结构

1 个答案: